Question

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients. Are x2-7x+12

Answer 1

Working out:

We have,

• f(x) = x² - 7x + 12

Factorising by using middle term factorisation,

⇛ x² - 7x + 12

⇛ x² - 3x - 4x + 12

⇛ x(x - 3) - 4(x - 3)

⇛ (x - 3)(x - 4)

The zeroes of f(x) are given by f(x)

So, let's equate them to get the zeroes of f(x),

Now, f(x) = 0

$\sf{ \longrightarrow{ {x}^{2} - 7x + 12 = 0}}$

$\sf{ \longrightarrow{(x - 3)(x - 4) = 0}}$

$\sf{ \longrightarrow{x - 3 = 0 \: or \: x - 4 = 0}}$

$\sf{ \longrightarrow{x = 3 \: or \: x = 4}}$

Hence, the zeroes of f(x) are 3 and 4

Now, verifying the relationship of the zeroes with the coefficients of the terms.

❒ Sum of zeroes = 3 + 4 = 7

$\sf{ \longrightarrow{ - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } = - \dfrac{ - 7}{1} = 7}}$

$\therefore$Sum of 0s = -coeff. of x / -coeff. of x²

❒ Product of zeroes = 3 × 4 = 12

$\sf{ \longrightarrow{ \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } = \dfrac{12}{1} = 12}}$

$\therefore$Product of 0s= cons. term / coeff. of x²

Hence, verified !!

Answer 2

Answer:

Mark as brainliest!!!

Step-by-step explanation:

given polynomial x²-7x+12

x²-4x-3x+12

x(x-4)-3(x-4)

(x-3)(x-4)

3,4 are the zeros

sum of zeros=7=-b/a=7

product of zeros =12=c/a=12

hence verified.

I hope this will help u ;)

OR,

Hai friend,

here is your answer,

The polynomial is x²-7x+12

a=1,b=-7,c=12

Factors of x²-7x+12

=x²-4x-3x+12

=x(x-4)-3(x-4)

=(x-4)(x-3)

Zeroes of the polynomial

x²-7x+12 =0

(x-4)(x-3)=0

x= 4,3

ZEROES = 4, 3

SUM OF ROOTS = 4+3=-b/a=7/1

=> 7=7

PRODUCT OF ROOTS= 4*3=c/a

=> 12=12